User talk:Xantharius

Course structure
Created the article coarse structure today: please help with formatting and so on if you like. Xantharius 20:11, 1 May 2006 (UTC)

Clout Fantasy
Hi, can we talk about how to get the "non-neutral content marker" off the Clout Fantasy page? I've re-edited it several times to attempt to make it neutral and would like to know if I've succeeded. I"m not sure how to proceed from here. Thanks, Jirel.

Generational secrets
Thanks dear! Might have to slug a couple of whiskey neat, after this little battle over Generational secrets. I think I've bent a few minds. To the "Revelator" and your Winged Self, Pax et Agape, Tatee Estelle

Joe Bray
Thanks for the tip. I must have forgotten to plug that in there. Joeraybray 03:06, 18 July 2007 (UTC)

Um...
My edit said that Reals were not closed, so it's not clear why you reverted it.- (User) WolfKeeper (Talk) 01:40, 2 March 2008 (UTC)

Thanks for converting math formulas to HTML
However:

-- Dominus (talk) 03:40, 16 March 2008 (UTC)

Zero-product property
I saw your edit, but isn't it too obvious to be worth mentioning? By definition, an integral domain is a commutative ring with unity without zero divisors. Saying that every (commutative rings with unity that is not an integral domain) has zero divisors is almost rephrasing the definition. If we drop the unity, then it's easy to construct a commutative ring without zero divisors that is not an integral domain: take, for example, the subring generated by x in the ring Z[x]. (I brag myself as The Overlord of Counter-examples :-) ). Albmont (talk) 12:47, 25 March 2008 (UTC)


 * Until a few days ago the article said that a ring which was not an integral domain had zero divisors, which is obviously untrue as the counter-example of the quaternions showed. I agree with you entirely that this is rephrasing the definition. But how else do we talk about rings which are not integral domains? Since the zero-product property is in fact the defining property of an integral domain, if we remove this then zero divisors must arise. I can't at the moment think of a better way of phrasing this which still highlights what happens when commutative rings fail to have the zero-product property. Xantharius (talk) 18:12, 25 March 2008 (UTC)


 * In fact, the integral domain is a ring with 3 independent axioms added to it: unit, commutativity and no zero divisors. And I just showed a commutative ring that is not an integral domain, that has no zero divisors: the ring of x p(x), where p(x) is a polynomial with integer coefficients. BTW, I think I can provide a quite pathological example of a commutative ring that is not an integral domain, has no zero divisors and can't be immersed into an integral domain Albmont (talk) 18:30, 25 March 2008 (UTC)


 * Well, it all depends on one's starting point (whether we are discussing commutative rings in the first place, for example) to determine what makes an integral domain an integral domain, but from the perspective of the zero-product property neither unity nor commutativity are the real focus: it's the zero-product property that counts, in my view, and I wonder whether it benefits the reader to make too much of a deal about this. Still, correctness is desirable.


 * Of course, any commutative ring R with unity with a rule for multiplication defined by ab = 0 for all a, b in R is an example of a commutative ring with unity which is not an integral domain, because everything is a zero divisor. But what's your example? Xantharius (talk) 19:47, 25 March 2008 (UTC)


 * Your example is not a commutative ring with unity: there's no unity, because 1.a = 0. If a commutative ring with no zero divisors can be expanded to an integral domain, then it can be immersed in the field of quotients of that domain, so this suggests the construction of the pathological example: it's enough to have one where $$x^n = a\,$$ has n+1 (or more) solutions. But I can't find a simple enough counter-example. Albmont (talk) 20:16, 25 March 2008 (UTC)


 * Oops. I think I can't build such counterexample. I am almost convincing myself of the opposite: every commutative ring with the zero-product property can be immersed in a field. Too bad, that's one class of counterexamples that vanish :-( Of course if x^2 = a has three distinct roots, then there are zero divisors. It's quite easy to prove it for x^3 = a, and probably there's a trivial induction to extend to n. Albmont (talk) 20:29, 25 March 2008 (UTC)

Have a cookie


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Oxford Wikimania 2010 and Wikimedia UK v2.0 Notice
Hi,

As a regularly contributing UK Wikipedian, we were wondering if you wanted to contribute to the Oxford bid to host the 2010 Wikimania conference. Please see here for details of how to get involved, we need all the help we can get if we are to put in a compelling bid.

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We look forward to hearing from you soon, and we send our apologies for this automated intrusion onto your talk page!

Addbot (talk) 07:50, 31 August 2008 (UTC)

Intermediate value theorem
The proof seems to be missing some formulas. Katzmik (talk) 12:23, 2 September 2008 (UTC)

D&D articles for Wikipedia 0.7
Hi there! :)

As someone who's worked on D&D and/or RPG articles before, I'm inviting you to participate in our goal to both improve articles that have been selected to be placed in the next Wikipedia DVD release, as well as nominate more to be selected for this project. Please see the WikiProject D&D talk page for more details. :) BOZ (talk) 18:33, 24 September 2008 (UTC)

File:Zerodivisor.png listed for deletion
A file that you uploaded or altered, File:Zerodivisor.png, has been listed at Files for deletion. Please see the to see why this is (you may have to search for the title of the image to find its entry), if you are interested in it not being deleted. Thank you. Skier Dude ( talk ) 04:22, 30 October 2011 (UTC)

ArbCom elections are now open!
MediaWiki message delivery (talk) 13:52, 23 November 2015 (UTC)

Women in Red World Contest
Hi. We're into the last five days of the Women in Red World Contest. There's a new bonus prize of $200 worth of books of your choice to win for creating the most new women biographies between 0:00 on the 26th and 23:59 on 30th November. If you've been contributing to the contest, thank you for your support, we've produced over 2000 articles. If you haven't contributed yet, we would appreciate you taking the time to add entries to our articles achievements list by the end of the month. Thank you, and if participating, good luck with the finale!